Specialist Maths — Question 17

QCAA 2020, Paper 1 · 7 marks

Q17 · 2020 · Technology-freeComplex familiar7 marks

QUESTION 17 (7 marks)

Determine the smallest positive value of

a

given

\int_{-a}^{a}\left[1+\left(\frac{\sec(2x)+\tan(2x)}{\csc(2x)+1}\right)^2\right]dx=1.

WORKED SOLUTION

7 marks

ANSWER

a=\frac{\pi}{8}

Worked solution

The fraction simplifies to

\tan(2x)

, so the integrand is

1+\tan^2(2x)=\sec^2(2x)

. Hence

\int_{-a}^{a}\sec^2(2x)\,dx=\frac12[\tan(2x)]_{-a}^{a}=\tan(2a)=1.

For the smallest positive solution,

2a=\pi/4

, giving

a=\pi/8

correctly expresses the expression inside the brackets of the integrand in terms of

\sin(2x)

and

\cos(2x)

[1 mark]

correctly simplifies the expression inside the brackets of the integrand

[1 mark]

uses a suitable Pythagorean identity to express the integrand as a single trigonometric expression

[1 mark]

determines the definite integral equation

[1 mark]

substitutes limits of integration

[1 mark]

solves equation to determine the smallest positive value of

a

[1 mark]

shows logical organisation communicating key steps

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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